Thread: Understanding properties of continuous functions w/ intervals

Hello all!

I am working on a written (essay) assignment regarding the properties of continuous functions on a closed interval [a,b]. In my case, I chose to discuss this set of theorems, but I am having trouble understanding one in specific. Graphically I can follow it along. I believe I have a pretty good grasp on Theorem 1 (f(a) < 0 < f(b) means we must have an f(x) = 0), Theorems 4 and 5 (if f(a) < c < f(b), then we must have f(x) = c. Theorem 5 is the same but with f(a) > c > f(b)).

The theorem I am having trouble understanding has to do with what are called Theorem 10/11. Specifcally, I am looking at Theorem 11, which is an extension of theorem 10. Here is Theorem 10:



This is more the part I am having trouble understanding. From the graphical analysis, it seems like it's manipulating an interval (or choosing an interval) such that when we look for f(y) <= f(x), , that f(y) <= f(x) for the entire line, not just at a point (because there could be a function where y0 isn't necessarily the lowest point on the entire line, just on an interval) . Is that a correct way of describing what theorem 10 entails?

Now, here is theorem 11:


To me, Theorem 11 is saying something that I can graphically show easily, but am having trouble describing just in plain English. Seems just to show that up to a certain point, if we take the function of a number below a constant c, that we will be guaranteed to have a solution to the equation and if c < m, it fails.

Is there a better way to describe Theorem 10 and 11? As always, thanks again in advance, you guys are the best!

Greetings,

You're lucky. I've never had to write any paper for my math class (those that's just my naivety--I'm still in high school).

Differentiation of the equation in (10) yields a polynomial of odd (n - 1) degree. From the fundamental theorem of algebra (or some corollary of it), we realize that a polynomial of odd degree must have at least one real root, and thus the original function must have an absolute minimum (as the coefficient to is positive) along . For theorem 11, imagine the left hand side instead being set to f(x). Now, f(x) doesn't necessarily have to have any real solutions, as it is even. Concerning c in this function, the polynomial is shifted down c units. As f(x) "points up" (as , and as , ), and if it has no real solutions, then the x-axis lies below the absolute minimum of f(x). Theorem (11) says that there exists a numberm such that f(x) is moved beyond the x-axis, giving it a real solution. (It has to have an even number of real solutions and they can be multiplicative.) If c is greater than or equal to this number m, then it has been shifted down the appropriate number of units to attain real solutions, but if c is less than this number m, then we haven't real solutions.

edit: I sorta implied it with "as the coefficient to is positive", but that and the remark about f(x) "pointing up" are directly related to one another. The sign would determine with way the polynomial "points". Also, there exists no loss of generality in the assumption that f(x) = c doesn't already have real solutions (hence the absolute minimum lies on or below the x-axis), for we're looking for a number m that satisfies the inequalities.